# Thread: Area of a region

1. ## Area of a region

Hey all,

so i want the area in grey like the image shows, calculating it with a double integral, in polar coordinates.

So for the lower half i put:

r = [-2sin theta , 2cos theta]
theta = [-pi/4 , 0]

for the upper half its almost the same.

r = [2sin theta , 2cos theta]
theta = [0 , pi/4]

Then i set up the integral with those limits and r dr dtheta.

The solution according to the person who made the exercise is 2. But i somehow can't get that. Is there anything wrong here or just some mistake on the calculation of the integrals?

Thanks

EDIT: my upper half's result is 1, so its the lower one that is screwed, unless this is some kind of coincidence.

EDIT2: Sorry to bother. I found the mistake. It was indeed on the lower half. "r" can't take negative values, so it takes the same values as the upper limit. Only the range of theta is different.

Thanks! Maybe this will help anyone anyway.

2. Ok, back here again =)

How to put the grey area in polar coordinates, to use as a double integral?

Theta will go [0, pi/4]

Then there should be two functions of "r", one for the left and one for the right side of the area. I'm not managing to find those.

3. It isn't hard. You just need to imagine some triangles.

You know that tan(theta) = y/x and r = sqrt(x²+y²). You know x (1 for the left side, 2 for the right side).

You can then express y as x * tan(theta), so r = sqrt(x² + (x*tan(theta))²) -> voilà

4. Thank you pedro, but I'm not sure that will do it though, since for the integral we can't have any x or y. R should only be a function of theta.

hmm

5. Ok, i found the solution. It's actually very simple.

The larger triangle is limited laterally by x = 2, the smaller one by x = 1.
x = rcostheta

so rcostheta = 2

this gives, r = 2 / costheta

(To simplify, 1/ costheta = sectheta)

So the values for each variable are

Theta [0, pi / 4]
R [ sectheta, 2sectheta]